递归——递归数据结构相关问题解决—以American Heritage问题为例
问题描述:
Farmer John takes the heritage of his cows very seriously. He is not, however, a truly fine bookkeeper. He keeps his cow genealogies as binary trees and, instead of writing them in graphic form, he records them in the more linear
tree in-order" and
tree pre-order" notations.
Your job is to create the `tree post-order" notation of a cow"s heritage after being given the in-order and pre-order notations. Each cow name is encoded as a unique letter. (You may already know that you can frequently reconstruct a tree from any two of the ordered traversals.) Obviously, the trees will have no more than 26 nodes.
Here is a graphical representation of the tree used in the sample input and output:
C / / B G / / A D H / E F The in-order traversal of this tree prints the left sub-tree, the root, and the right sub-tree. The pre-order traversal of this tree prints the root, the left sub-tree, and the right sub-tree. The post-order traversal of this tree print the left sub-tree, the right sub-tree, and the root. --------------------------------------------------------------------------------
题目大意:
给出一棵二叉树的前序遍历 (preorder) 和中序遍历 (inorder),求它的后序遍历 (postorder)。
你需要知道的:
1:二叉树的 相关定义可以在书上或者网上找到。
2:样例 输入输出反映的二叉树在上面。
输入描述
Line 1:
The in-order representation of a tree.
Line 2:
The pre-order representation of that same tree.
输出描述
A single line with the post-order representation of the tree.
样例输入
ABEDFCHG
CBADEFGH
样例输出
AEFDBHGC
解题思路:
这是一个关于递归数据结构的算法题。具体来说是个二叉树,无论多大的二叉树都可以看成是一个根结点和左右孩子的结合。
所以只要关注三个点就够了。
根据前序遍历的序列我们可以知道第一个元素为树的根结点,再更具中序遍历的序列,可以由已知的根结点来找到左右孩子。如果左右孩子也可看做是树(即不为空),那么继续重复以上过程。
对三个结点的树来说,根节点是最后输出,所以应该采用头递归,将输出结点的步骤放在最后。
源码如下:
#include<stdio.h> #include<string.h> void hoge(char *pre,char *in,int length){ char c=pre[0]; if(length == 0){ return; } if(length ==1){ printf("%c",c); return; } int i=0; while(in[i]!=c){ i++; } hoge(pre+1,in,i); hoge(pre+i+1,in+i+1,length-1-i); printf("%c",c); } int main(){ char pre[30],in[30]; int length; while(scanf("%s%s",pre,in)!=EOF){ length = strlen(pre); hoge(in,pre,length); } return 0; }下面还有一种不太理想却有助于理解的思路
参考如下
#include<stdio.h> #include<malloc.h> #include<string.h> typedef struct BTNode{ char data; struct BTNode *lchild; struct BTNode *rchild; }BTNode; //创建二叉树结点 BTNode *CreateBTree(char a[],char b[],int n){ int k; if(n<=0) return NULL; //递归出口:没有剩余结点时,返回空结点 char root = a[0] ; BTNode *bt = (BTNode*)malloc(sizeof(BTNode)); //创建每个分结点的根结点 bt->data = root; for(k=0;k<n;k++) if(b[k] == root) break; bt->lchild = CreateBTree(a+1,b,k); bt->rchild = CreateBTree(a+k+1,b+k+1,n-k-1);//递归体:确定了根节点,连接左右结点 return bt; } //时间复杂度为log(N) void LRD(BTNode *bt){ if(bt == NULL) return;//递归出口:遇到空结点时则弹出栈 LRD(bt->lchild);//递归体:对一棵树先完成输出左孩子再完成输出右孩子最后输出自己 LRD(bt->rchild); printf("%c",bt->data); }//时间复杂度为N int main(){ char a[10000] = ""; char b[10000] = ""; while(scanf("%s%s",a,b)!=EOF){ int length = strlen(a); BTNode *bt = (BTNode*)malloc(sizeof(BTNode)); bt=CreateBTree(b,a,length); LRD(bt); } return 0; }感悟:
对于递归结构体的题目,可以根据其定义将很多细节整体化,化繁为简。